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NEET CHEMISTRYMedium

The correct shape and hybridisation for XeF4\text{XeF}_4 are:

A

octahedral, sp3d2sp^3d^2

B

trigonal bipyramidal, sp3d2sp^3d^2

C

planar triangle, sp3d3sp^3d^3

D

square planar, sp3d2sp^3d^2

Step-by-Step Solution

In XeF4\text{XeF}_4, the central Xenon (Xe) atom has 8 valence electrons. It forms 4 single bonds with 4 Fluorine (F) atoms using 4 of its valence electrons. The remaining 4 valence electrons form 2 lone pairs. The steric number (total electron domains) = Number of bond pairs (4) + Number of lone pairs (2) = 6. A steric number of 6 corresponds to sp3d2sp^3d^2 hybridisation. According to VSEPR theory, the 6 electron domains arrange themselves in an octahedral geometry to minimize repulsion. To minimize the strong lone pair-lone pair repulsions, the two lone pairs occupy opposite axial positions. This leaves the 4 Fluorine atoms in the equatorial positions, resulting in a square planar molecular shape.

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